I am trying to get that the path integral of the parallel transport action is the Wilson loop. Here is the setting:

Let $w$ be a complex vector dimension $N$, and $A_{\mu}$ a fixed Yang-Mills connection. We will work with $G=SU(N)$. Using the parallel transport equation and the constraint:

$$i\frac{dw}{d\tau} = \frac{dx^{\mu}}{d\tau} A_{\mu}w$$ $$w^{\dagger}w=1 $$ We construct the following action with a Lagrange multiplier $\lambda$ enforcing the constraint above. Evidently the equation of motion of this action is the parallel transport equation. $$S_w = \int \left(i w^{\dagger}\frac{dw}{dt} + \lambda(w^{\dagger}w-1)+w^{\dagger}A(x(\tau))w \right)d\tau$$

This vector satisfies: $$[w_i,w_j^{\dagger}]=\delta_{ij}$$

Now let $\tau \in \mathbf{S}$ to allow for large gauge transformations. I finally arrive to the following path integral, where I had to insert the $w_i$ factors for it to not vanish:

$$ Z_w[A]= \int e^{iS_w (w;\lambda;A)} w_i(\tau=\infty)w_i^{\dagger}(\tau=-\infty) \mathcal{D}\lambda\mathcal{D}w \mathcal{D}w^{\dagger}$$

I am supposed to get that

$$Z_w[A] = tr \mathcal{P}e^{i\int A d\tau}$$

How do I compute that specific path integral?

  • $\begingroup$ I think, that commutation equation, that you wrote, are in canonical quantization. But you are interested in path integral quantization of such system, right? $\endgroup$
    – Nikita
    Apr 6, 2020 at 16:22
  • $\begingroup$ Yes. I was providing background for both quantizations, but mainly I'm interested in path integral. $\endgroup$ Apr 6, 2020 at 17:09
  • $\begingroup$ Is it obvious, that path integral vanishes, if one didn't inserted $w_i$ factors? $\endgroup$
    – Nikita
    Apr 6, 2020 at 19:21

1 Answer 1


I will follow Monopoles and Wilson Lines, by David Tong, Kenny Wong.

Let's work in perturbation theory. Action: $$ S_w = \int \left(i w^{\dagger}\frac{dw}{dt} + \alpha(w^{\dagger}w-\kappa)+w^{\dagger}A(x(\tau))w \right)d\tau $$ Propogator for free (non-interacting) part of action, in following we use $\langle \dots \rangle$ for averaging by free theory: $$ \langle w_i^\dagger(\tau_1) w_j(\tau_2)\rangle = \theta(\tau_1-\tau_2) \delta_{ij} $$

  1. Vacuum bubbles

As usual in perturbation theory, we can factorise vacuum diagrams. You will obtain up to notations:

All $n ≥ 2$ factors vanish because the product of the step functions vanishes everywhere except on a set of measure zero. So we have only one contribution ($\theta(0)=1/2$): $$ \exp\left(i (N/2 - \kappa )\int dt\; \alpha(t)\right) = \exp\left(-i \kappa_{eff} \int dt\; \alpha(t)\right) $$

  1. Path integral with insertions

$$ Z_w[A]= \int\mathcal{D}\lambda\mathcal{D}w \mathcal{D}w^{\dagger}\; e^{iS_w (w;\lambda;A)} w_i(\tau=\infty)w_i^{\dagger}(\tau=-\infty) $$

This integral correspond to following series of diagrams (times to vacuum bubbles factor):

This series correspondence to:

Including vacuum bubbles we left with:

$$ Z_\omega[A] = W[A]\int D\alpha e^{-i\int dt \;\alpha(t) (\kappa_{eff}-1)} $$

If $\kappa_{eff}= 1$, we obtain:

$$ Z_\omega[A] = W[A] $$

  • $\begingroup$ Could you summarize why without insertion of the $w$ and $w^{\dagger}$ the path integral vanishes? Also, what would be the difference if there is no $\lambda$ apart from not conserving the norm of $w$? $\endgroup$ Apr 7, 2020 at 16:42
  • $\begingroup$ Without insertion you will have only $\int D\alpha e^{-i\int dt \alpha(t)\kappa_{eff}}= \delta(\kappa_{eff})$. $\endgroup$
    – Nikita
    Apr 7, 2020 at 18:58
  • $\begingroup$ What about without imposing $\alpha$? $\endgroup$ Apr 7, 2020 at 19:08
  • $\begingroup$ I don't understand your question. $\alpha$ is present in action from initial step. $\endgroup$
    – Nikita
    Apr 7, 2020 at 20:26
  • 1
    $\begingroup$ Yes, formally this integral is infinity, so you need be ore careful when deal with correlators. You need divide integral by average of identity operator. Infinity will be cancel. $\endgroup$
    – Nikita
    Apr 8, 2020 at 1:42

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